What is the solution for 2x^{''}+4x+20x=20\H(t) ?

The solution for 2x^{''}+4x+20x=20\H(t) is

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2x^{''}+4x+20x=20\H(t)

2 x^{^{''}} + 4 x + 20 x = 20 H (t)

x = c_{1} cos (23 t) + c_{2} sin (23 t) + cos (23 t) \cdot \int - \frac{5 3 sin ( 2 3 t ) H ( t )}{3} d t + sin (23 t) \cdot \int \frac{5 3 cos ( 2 3 t ) H ( t )}{3} d t

Solution steps

2 x^{''} (t) + 4 x + 20 x = 20 H (t)

Solve linear ODE:

x = c_{1} cos (23 t) + c_{2} sin (23 t) + cos (23 t) \cdot \int - \frac{5 3 sin ( 2 3 t ) H ( t )}{3} d t + sin (23 t) \cdot \int \frac{5 3 cos ( 2 3 t ) H ( t )}{3} d t

x = c_{1} cos (23 t) + c_{2} sin (23 t) + cos (23 t) \cdot \int - \frac{5 3 sin ( 2 3 t ) H ( t )}{3} d t + sin (23 t) \cdot \int \frac{5 3 cos ( 2 3 t ) H ( t )}{3} d t

2 y^{^{''}} - y^{^{'}} - y = 25 e^{- x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 5 y = e^{x} sin (2 x)

y^{^{''}} - 4 y = 12 sin (2 t)

y^{^{''}} + 6 y^{^{'}} + 45 y = 6

2 y^{^{''}} + 3 y^{^{'}} + 5 y = 3 cos (5 t)

What is the solution for 2x^{''}+4x+20x=20\H(t) ?
The solution for 2x^{''}+4x+20x=20\H(t) is